A radioactive element decays by `beta-`emission. A detector racords `n`-beta particles in `2` sec and in next `2` sec it records `0.65n`-beta particle

Correct Answer - [T] `(c_(t))/(c_(0))=e^(-lambdax(1)/(lambda))=((1)/(e))=f_(1)=0.36` (left ) then decay `0.633(f_(1))` `(c_(1))/(c_(0))=e^(-lambdax(ln2)/(lambda))=2^(-1)=((1)/(2))=f_(2)=0.5` (left)decay `0.5(f_(2)` `f_(1) gt f_(2)`

Correct Answer - C `lamda=(lamda_(1)+lamda_(2))=(1)/(1620)+(1)/(520)=2.54xx10^(-3)yr^(-1)` `therefore" " t_(1//2)=(1n(2))/(lamda)=272.8 yr` (1/4) th of the material remainsintact after 2 half-lives.

Correct Answer - C We have, `lamda_(B)N_(B)=lamda_(A)N_(A)` `therefore" "N_(B)=(lamda_(A))/(lamda_(B)).N_(A)` The give time is equivalent to two half-lives of A. Hence, `N_(A)=(N_(0))/(4)" "(thereforeN=N_(0)((1)/(2))^(2)` `N_(B)=(lamda_(A))/(lamda_(B))((N_(0))/(4))`

Correct Answer - C Given in 1st two seconds `100beta` particles and in next two seconds `50beta` particles are emitted. `:.T_(1//2)=2s` Hence, mean life, `T=(t_(1//2))/(0.693)=(2)/(0.693)`

Correct Answer - C We have, `lamda=(0.693)/(T_(1//2)` Mean life, `tau=(1)/(lamda)=(t_(1//2))/(0.693)=(2)/(2.303log_(2))=(2)/(In2)s`

Correct Answer - A Due to emission os `3alpha` particles, the mass number is reduced by 12 while atomic number is decresed by 6. Due to emission of `beta` particle the...

The process for negative beta decay is given by equation `._(19)^(40)K_(21) rarr ._(20)^(40)Ca_(20) +e^(-) + barv` and the Q value is found from equation using atomic masses : `Q_(beta) [m(.^(40)K)...

To determine the Q value for this decay, we first need to find the mass of the product nucleus `.^(12)C` in its excited state. In the ground state, `.^(12)C` has...

Correct Answer - isotope

Correct Answer - `zM^(A-4) `